# The prime number theorem with the classical error term

*Written by GPT-6.1 Sol (OpenAI) in Codex, Ultra setting, October 2026. Self-checked by the writing AI. Public domain (CC0).*

The prime number theorem says that the weighted prime count is asymptotic to $x$. Its error depends on how far zeros stay from the line one. The logarithmic zero-free region gives a saving exponential in $\sqrt{\log x}$. We prove that error first for prime powers, then transfer it to primes. A contour for the reciprocal of zeta gives the corresponding cancellation among Möbius values.

We use the zero-free region and reciprocal bounds proved in Nonvanishing on the line one and a zero-free region, Theorems 3.2 and 4.1; the sharp explicit formula and finite Perron formula from Perron's formula and the explicit formula for prime counting, Theorem 2.2 and Proposition 1.2; and the reciprocal-ordinate estimate in Counting the zeros, Theorem 4.1. Elementary prime-power comparisons and partial summation are proved in Counting primes by elementary means, Theorems 2.1 and 5.2. The ordinary convergence $\sum\mu(n)/n=0$ is proved in The prime number theorem, Theorem 4.1 and Solution 3. The same results are treated in Koukoulopoulos’s preliminary version, Chapter 8, Theorem 8.1 and its proof, listed below.

Throughout, $\psi(x)=\sum_{n\le x}\Lambda(n)$, $\theta(x)=\sum_{p\le x}\log p$, and $M(x)=\sum_{n\le x}\mu(n)$. The letter $c$ denotes a positive absolute constant whose value may decrease between estimates. We retain explicit names for the constants in the zero-free region when using them in a contour. As before, $\operatorname{li}(x)=\operatorname{Ei}(\log x)$ and $\operatorname{Li}(x)=\int_2^xdt/\log t$ differ by the constant $\operatorname{li}(2)$.

The integration and complex-analysis tools used in this lesson are proved in Dirichlet series and Euler products, Appendix A, Lemmas A.1–A.4.

## 1. A quantitative prime-power count

**Theorem 1.1.** There is a positive absolute constant $c$ such that, for every $x\ge2$,
$$
\psi(x)=x+O\bigl(xe^{-c\sqrt{\log x}}\bigr).
\tag{1.1}
$$

**Proof.** The preceding zero-free proof supplied $C=1/2000$ with
$$
\beta<1-\frac C{\log(|\gamma|+2)}
$$
for every nontrivial zero $\rho=\beta+i\gamma$. For $|\gamma|<T$, $T\ge2$, this implies
$$
|x^\rho|\le x\exp\left(-\frac{C\log x}{\log(T+2)}\right).
$$
The zero-counting estimate and partial summation give $\sum_{|\gamma|<T}1/|\rho|\ll\log^2(T+4)$, including the finitely many small ordinates. Hence the sharp explicit formula gives
$$
|\psi(x)-x|\ll
x\log^2(T+4)e^{-C\log x/\log(T+2)}
+\frac{x\log^2(xT)}T+\log x.
\tag{1.2}
$$
The last term includes both the nearest-prime-power error, whose minimum factor is at most one, and the change from $\psi_0$ to $\psi$. The constant and trivial-zero terms are bounded for $x\ge2$ and are absorbed as well. Thus (1.2) holds at prime powers and between them, with no restriction on the chosen height.

Put $L=\log x$ and $T=e^{\sqrt L}$. For sufficiently large $L$, $\log(T+2)\le2\sqrt L$ and $\log(T+4)\ll\sqrt L$. The first term in (1.2) is $O(xL e^{-(C/2)\sqrt L})$, and the second is $O(xL^2e^{-\sqrt L})$. For each fixed $a>0$ and integer $m\ge0$, $L^m e^{-a\sqrt L}=O(e^{-(a/2)\sqrt L})$: set $v=\sqrt L$ and use the boundedness of $v^{2m}e^{-(a/2)v}$. The remaining $\log x$ is also absorbed in $xe^{-c\sqrt L}$. These observations prove (1.1) for sufficiently large $x$. Enlarging the implied constant covers the fixed interval below that threshold. $\square$

This is a proof with an absolute positive saving. It does not claim a useful numerical error at moderate $x$ from our deliberately small zero-free constant.

## 2. Passing from powers to primes

**Theorem 2.1.** For some positive absolute constant $c$ and all $x\ge2$,
$$
\theta(x)=x+O\bigl(xe^{-c\sqrt{\log x}}\bigr),\qquad
\pi(x)=\operatorname{li}(x)+O\bigl(xe^{-c\sqrt{\log x}}\bigr).
\tag{2.1}
$$

**Proof.** The proved elementary comparison $0\le\psi(x)-\theta(x)\ll\sqrt x$ gives the first result, since $\sqrt x=O(xe^{-c\sqrt{\log x}})$ for each fixed $c>0$. Write $\theta(t)=t+E(t)$, with $E(t)\ll te^{-a\sqrt{\log t}}$ for some $a>0$. Partial summation, retaining the lower endpoint, gives
$$
\pi(x)=\frac{\theta(x)}{\log x}
+\int_2^x\frac{\theta(t)}{t\log^2t}\,dt.
$$
The terms with $\theta(t)$ replaced by $t$ equal
$$
\frac{x}{\log x}+\int_2^x\frac{dt}{\log^2t}
=\operatorname{Li}(x)+\frac2{\log2}.
$$
This identity follows by differentiating $t/\log t$. Its constant difference from $\operatorname{li}(x)$ is harmless for the present error.

The endpoint error is at most $Cxe^{-a\sqrt{\log x}}/\log x$. For large $x$, split the error integral at $\sqrt x$. On the first portion, the elementary bound $|E(t)|\ll t$ gives $O(\sqrt x)$. On the second portion, $\sqrt{\log t}\ge\sqrt{\log x}/\sqrt2$; therefore its modulus is at most
$$
Ce^{-(a/\sqrt2)\sqrt{\log x}}
\int_{\sqrt x}^x\frac{dt}{\log^2t}
\ll xe^{-(a/\sqrt2)\sqrt{\log x}}.
$$
The constant term and the initial interval are absorbed by reducing $c$ and enlarging its implied constant. This proves the second assertion. $\square$

## 3. A contour for the Möbius sum

The reciprocal is holomorphic at one and has a zero there. Its counting contour consequently has no main-term residue.

**Lemma 3.1 (a finite formula for bounded coefficients).** Suppose $|b(n)|\le1$ and $D(s)=\sum b(n)n^{-s}$ for $\Re s>1$. If $X=N+1/2\ge2$ with integer $N$, $a_X=1+1/\log X$, and $T\ge2$, then
$$
\sum_{n\le N}b(n)
=\frac1{2\pi i}\int_{a_X-iT}^{a_X+iT}D(s)\frac{X^s}s\,ds
+O\left(\frac{X\log X}T\right).
\tag{3.1}
$$

**Proof.** Apply the finite Perron formula. There is no term $n=X$ because $X$ is a half integer. For $n\le X/2$ or $n\ge2X$, $|\log(X/n)|\ge\log2$, so the error is at most $CX^a\sum n^{-a}/T\ll X\log X/T$, using $X^a=eX$ and $\zeta(a)\le1+1/(a-1)$. In the middle range, $(X/n)^a\ll1$ and $|\log(X/n)|\ge|X-n|/(2X)$. Every distance is at least $1/2$. The two terms at distances below one cost $O(X/T)$, because $\min(1,CX/T)\le CX/T$. Grouping all remaining distances in unit intervals gives at most two integers per group and the bound $CX\sum_{j\le2X}1/(Tj)\ll X\log X/T$. This proves (3.1) directly, without an endpoint convention for signed coefficients. $\square$

**Theorem 3.2.** There is a positive absolute constant $c$ such that
$$
M(x)=O\bigl(xe^{-c\sqrt{\log x}}\bigr)\qquad(x\ge2).
\tag{3.2}
$$

**Proof.** The Euler product proves $D(s)=1/\zeta(s)=\sum\mu(n)n^{-s}$ for $\Re s>1$. Take $X=\lfloor x\rfloor+1/2$, so $M(X)=M(x)$, and use Lemma 3.1. Put
$$
d=\frac1{10000},\qquad \kappa=\frac d4,\qquad
\eta=\frac\kappa{\log(T+4)},\qquad b=1-\eta.
$$
The rectangle $b\le\sigma\le a_X$, $|t|\le T$ is zero-free by lesson eight, Theorem 3.2. For $2\le|t|\le T$, its left edge satisfies
$$
\eta\le\frac d{2\log(|t|+2)},
$$
so Theorem 4.1 of that lesson gives $|D(s)|\ll\log(T+4)$ throughout this part of the rectangle. The portion $|t|\le2$ lies in the fixed compact rectangle
$$
1-\frac\kappa{\log6}\le\sigma\le1+\frac1{\log2},\qquad |t|\le2.
$$
It is zero-free, and $D$ extends holomorphically at the pole $s=1$ with value zero. Continuity on that compact set therefore bounds $D$ by an absolute constant there. Together these facts give the required bound on every side, including the small-height part.

Cauchy's theorem shifts the finite integral in (3.1) to $\sigma=b$; no pole is crossed, and $b>3/4$ keeps the kernel pole zero outside the rectangle. The new vertical integral is bounded by
$$
CX^b\log(T+4)\int_{-T}^T\frac{dt}{|b+it|}
\ll X^{1-\eta}\log^2(T+4).
$$
Each horizontal integral is at most
$$
\frac{C\log(T+4)}T\int_b^{a_X}X^\sigma d\sigma
\ll\frac{X\log(T+4)}T.
$$
The interval has bounded length and $X^{a_X}=eX$, which suffices for the last inequality. Consequently
$$
|M(x)|\ll
Xe^{-\kappa\log X/\log(T+4)}\log^2(T+4)
+\frac{X(\log X+\log(T+4))}T.
\tag{3.3}
$$
Choose $T=e^{\sqrt{\log X}}$. Exactly the exponential absorption used in Theorem 1.1 gives $M(x)\ll Xe^{-c\sqrt{\log X}}$. Since $X=x+O(1)$ and $X$ is comparable to $x$, this is (3.2); a fixed initial interval is covered by changing the implied constant. $\square$

**Corollary 3.3.** For a positive absolute constant $c$,
$$
\sum_{n\le x}\frac{\mu(n)}n
=O\bigl(e^{-c\sqrt{\log x}}\bigr)\qquad(x\ge2).
\tag{3.4}
$$

**Proof.** Partial summation gives $\sum_{n\le x}\mu(n)/n=M(x)/x+\int_1^xM(t)t^{-2}dt$. Theorem 3.2 makes the infinite integral absolutely convergent, and $M(x)/x\to0$. The already proved ordinary limit $\sum\mu(n)/n=0$ therefore identifies that integral's value as zero. Thus the exact tail identity is
$$
\sum_{n\le x}\frac{\mu(n)}n
=\frac{M(x)}x-\int_x^\infty\frac{M(t)}{t^2}\,dt.
\tag{3.5}
$$
For $a>0$, substituting $v=\sqrt{\log t}$ computes
$$
\int_x^\infty e^{-a\sqrt{\log t}}\frac{dt}t
=2e^{-a\sqrt{\log x}}
\left(\frac{\sqrt{\log x}}a+\frac1{a^2}\right).
\tag{3.6}
$$
Apply this to the bound for $M(t)/t$ and absorb the factor $1+\sqrt{\log x}$ by decreasing the exponential constant. This proves (3.4). $\square$

The zero limit in (3.5) matters: a bound for $M$ and absolute convergence of the integral alone would leave its limiting constant unidentified. Our preceding proof fixes that constant before estimating the tail.

## 4. The scale of the cutoff

**Example 4.1.** Write $u=\log T$ and $L=\log x$. For $T\ge2$, $\log(T+2)\le2u$ and $\log(T+4)\ll u$. With $a=C/2=1/4000$, the two principal relative bounds in (1.2) have the forms
$$
E_{\mathrm{zeros}}=u^2e^{-aL/u},\qquad
E_{\mathrm{contour}}=(L+u)^2e^{-u},
\tag{4.1}
$$
up to absolute multiplicative constants. For $L+u\ge2$, increasing $u$ makes the contour term nonincreasing while enlarging the zero term. Indeed,
$$
\frac{d}{du}\log E_{\mathrm{contour}}=\frac2{L+u}-1\le0,
\qquad
\frac{d}{du}\log E_{\mathrm{zeros}}=\frac2u+\frac{aL}{u^2}>0.
$$ The simple choice $u=\sqrt L$ proves the required order even with our small $a$. Balancing the exponential factors alone gives $u=\sqrt{aL}$; including the polynomial factors shifts the balance by lower-order logarithmic terms.

![The two normalized error components against the logarithm of the cutoff height](figures/NT-ZETA-12/cutoff_balance.png)

*Figure 1. The two functions in (4.1), at the stated theoretical parameter $L=10^8$ and the proved constant $a=1/4000$. The vertical axis is logarithmic. Common implicit constants are suppressed; this plots the analytic bound's components, not a measured prime-counting error. The marked crossing includes both polynomial prefactors. The dotted line is the balance of the exponential factors alone. The Python source retains all parameters and reproduces the figure.*

## 5. Exercises and worked solutions

**Exercise 1.** Deduce the asserted estimate for $\pi(x)$ directly from (1.1), keeping the logarithmic-integral convention explicit.

**Solution 1.** First subtract the higher prime powers, whose total weight is $O(\sqrt x)$ by lesson two, to get the estimate for $\theta$. Then the exact partial-summation formula in Theorem 2.1 has main term $\operatorname{Li}(x)+2/\log2$. For its error, the range below $\sqrt x$ contributes $O(\sqrt x)$, and above $\sqrt x$ the exponential saving is at least $e^{-(a/\sqrt2)\sqrt{\log x}}$. The endpoint term has the same saving. Add the fixed difference $2/\log2-\operatorname{li}(2)$ to convert to $\operatorname{li}(x)$, and absorb it in the growing error envelope. This proves the claimed estimate with a possibly smaller positive constant.

**Exercise 2.** Prove the sharper Mertens estimate
$$
\sum_{p\le x}\frac1p=\log\log x+M_{\mathrm{pr}}+O\bigl(e^{-c\sqrt{\log x}}\bigr).
$$
Identify the constant with the one proved in lesson two.

**Solution 2.** Write $\pi(t)=\operatorname{li}(t)+E(t)$ using Theorem 2.1. Partial summation yields $\sum_{p\le x}1/p=\pi(x)/x+\int_2^x\pi(t)t^{-2}dt$. Differentiation of $\operatorname{li}(t)/t$ shows that its main terms equal
$$
\log\log x-\log\log2+\frac{\operatorname{li}(2)}2.
$$
Since $E(t)\ll te^{-a\sqrt{\log t}}$, the integral $\int_2^\infty E(t)t^{-2}dt$ converges absolutely by (3.6). Defining the constant by that integral leaves the error $E(x)/x-\int_x^\infty E(t)t^{-2}dt$. Formula (3.6) and exponential absorption bound it by $O(e^{-c\sqrt{\log x}})$. Its limiting constant must equal the constant already identified in lesson two, Theorems 3.1 and 4.2, because both are limits of $\sum_{p\le x}1/p-\log\log x$. Therefore
$$
M_{\mathrm{pr}}=\gamma+\sum_p\bigl(\log(1-1/p)+1/p\bigr).
$$
The notation $M_{\mathrm{pr}}$ keeps this constant distinct from the summatory function $M(x)$.

**Exercise 3.** Prove (3.4) from the bound for the Möbius sum and the preceding lesson's zero limit.

**Solution 3.** Partial summation followed by the zero limit gives exactly (3.5). Its endpoint is $O(e^{-a\sqrt{\log x}})$, and (3.6) bounds its tail by $O((1+\sqrt{\log x})e^{-a\sqrt{\log x}})$. The function $(1+v)e^{-(a/2)v}$ is bounded for $v\ge0$, so both terms are $O(e^{-(a/2)\sqrt{\log x}})$. This proves the assertion and explicitly controls the integral, rather than treating it as automatically the same size as the endpoint.

**Exercise 4.** Explain why optimization of these zero-free-region bounds alone cannot improve the square-root exponent scale. Distinguish this statement about a bound from a lower bound on the true arithmetic error.

**Solution 4.** Ignoring polynomial factors, the available normalized error envelope contains $e^{-u}$ and $e^{-aL/u}$. For every $u>0$,
$$
\min\{u,aL/u\}\le\sqrt{aL},\qquad
\max\{e^{-u},e^{-aL/u}\}\ge e^{-\sqrt{aL}}.
$$
If $u\le\sqrt{aL}$ the first inequality follows from its first entry; otherwise it follows from its second entry. Equality occurs at $u=\sqrt{aL}$. Thus optimizing the two available components cannot yield an exponent proportional to $L^\alpha$ with $\alpha>1/2$. The polynomial prefactors in (4.1) do not improve this obstruction and change the loss by logarithmic terms. This is a limitation of this pair of upper bounds and this contour method; it does not prove that $|\psi(x)-x|$ is as large as that envelope. A wider zero-free region supplies different components and can improve the exponent scale.

**Exercise 5.** For Liouville's function $\lambda(n)=(-1)^{\Omega(n)}$, prove $\sum_{n\le x}\lambda(n)=O(xe^{-c\sqrt{\log x}})$.

**Solution 5.** Complete multiplicativity gives the absolutely convergent Euler product
$$
\sum_{n\ge1}\frac{\lambda(n)}{n^s}
=\prod_p(1+p^{-s})^{-1}
=\frac{\zeta(2s)}{\zeta(s)}\qquad(\Re s>1),
$$
since $(1+z)^{-1}=(1-z)/(1-z^2)$. Absolute convergence follows from $|\lambda(n)|=1$ and $\sum n^{-\sigma}<\infty$; it justifies all these product identities. Apply Lemma 3.1 to these coefficients and shift the same rectangle as in Theorem 3.2. Its left edge has $b>3/4$, so the numerator is holomorphic throughout and $|\zeta(2s)|\le\zeta(2b)\le\zeta(3/2)$. In particular its pole at $s=1/2$ lies outside the rectangle. The denominator has no zero inside, and its pole at one produces a removable zero of the quotient. Thus the quotient is holomorphic, bounded by $C\log(T+4)$ on all sides. The same vertical and horizontal estimates give (3.3) for the Liouville sum. Choose $T=e^{\sqrt{\log X}}$, absorb the polynomial factors and replace $X=\lfloor x\rfloor+1/2$ by $x$ as before. This proves the required cancellation.

## Freely readable sources

- D. Koukoulopoulos, [*The Distribution of Prime Numbers*](https://dms.umontreal.ca/~koukoulo/documents/publications/primes.pdf), freely readable author preliminary version, 367 pages. Chapter 8, Theorem 8.1, with its proof on.